If 2tan A = 3tan B,then frac{sin 2B}{5 - cos | vedclass

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(A) Given $2	an A = 3	an B$,so $	an A = \frac{3}{2}	an B$. Let $	an B = t$,then $	an A = \frac{3}{2}t$. We know that $\sin 2B = \frac{2t}{1 + t^

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$	an (A - B)$

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(A) Given $2	an A = 3	an B$,so $	an A = \frac{3}{2}	an B$. Let $	an B = t$,then $	an A = \frac{3}{2}t$. We know that $\sin 2B = \frac{2t}{1 + t^2}$ and $\cos 2B = \frac{1 - t^2}{1 + t^2}$. Substituting these into the expression: $\frac{\sin 2B}{5 - \cos 2B} = \frac{\frac{2t}{1 + t^2}}{5 - \frac{1 - t^2}{1 + t^2}} = \frac{2t}{5(1 + t^2) - (1 - t^2)} = \frac{2t}{5 + 5t^2 - 1 + t^2} = \frac{2t}{4 + 6t^2} = \frac{t}{2 + 3t^2}$. Now,consider $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = \frac{\frac{3}{2}t - t}{1 + (\frac{3}{2}t)(t)} = \frac{\frac{1}{2}t}{1 + \frac{3}{2}t^2} = \frac{t}{2 + 3t^2}$. Thus,$\frac{\sin 2B}{5 - \cos 2B} = 	an(A - B)$.

If $2\tan A = 3\tan B$,then $\frac{\sin 2B}{5 - \cos 2B}$ is equal to

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